# Central Limit Theorem: $\mathbb{E}\left|\frac{1}{N} \sum_{i=1}^{N} X_i - \mu\right| = O\left(\frac{1}{\sqrt{N}}\right)$

I am self-studying the High Dimensional Probability book by Roman Vershynin and came across this problem:

Let $$X_1, X_2, \dots$$ be a sequence of i.i.d random variables with mean $$\mu$$ and finite variance. Show that

$$\mathbb{E}\left|\frac{1}{N} \sum_{i=1}^{N} X_i - \mu\right| = O\left(\frac{1}{\sqrt{N}}\right)$$ as $$N \rightarrow \infty$$

I feel I need to use central limit theorem somehow but not sure how to deal with absolute value

Notice that $$\mathbb{E}[|Z|]^2 \leq \mathbb{E}[Z^2]$$ for any random variable $$Z$$.note 1) Now by writing $$\bar{X}_N = \frac{1}{N}\sum_{i=1}^{N}X_i$$ and noting that $$\mathbb{E}[\bar{X}_N] = \mu$$, it follows that
\begin{align*} \mathbb{E}\big[\left|\bar{X}_N - \mu\right|\big]^2 &= \mathbb{E}\big[ \left|\bar{X}_N - \mathbb{E}[\bar{X}_N]\right| \big]^2 \leq \mathbb{E}\big[\left(\bar{X}_N - \mathbb{E}[\bar{X}_N] \right)^2 \big]\\ &= \mathbf{Var}\left( \bar{X}_N \right) = \frac{1}{N^2} \sum_{i=1}^{N} \mathbf{Var}(X_i) = \frac{\sigma^2}{N}, \end{align*}
where $$\sigma^2 = \mathbf{Var}(X_n)$$ is the common value of the variances of $$X_n$$'s. From this, we get
$$\mathbb{E}\left[\left|\frac{1}{N}\sum_{i=1}^{N}X_i - \mu\right|\right] \leq \frac{\sigma}{\sqrt{N}}$$
Note 1) If $$Z$$ has finite second moment, this inequality is equivalent to saying that $$\mathbf{Var}(|Z|) \geq 0$$. This fact itself can be proved in various ways such as Jensen's inequality, Cauchy-Schwarz inequality, etc.